# existence of the conditional expectation

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space  and $X$ be a random variable  . For any $\sigma$-algebra $\mathcal{G}\subseteq\mathcal{F}$, we show the existence of the conditional expectation $\mathbb{E}[X\mid\mathcal{G}]$. Although it is possible to do this using the Radon-Nikodym theorem  , a different approach is used here which relies on the completeness of the vector space  $L^{2}$. The defining property of the conditional expectation $Y=\mathbb{E}[X\mid\mathcal{G}]$ is

 $\mathbb{E}[1_{G}Y]=\mathbb{E}[1_{G}X]$ (1)

for sets $G\in\mathcal{G}$. We shall prove the existence of the conditional expectation for all nonnegative random variables and, more generally, whenever $\mathbb{E}[|X|\mid\mathcal{G}]$ is almost surely finite.

First, the conditional expectation of every square-integrable random variable exists.

###### Theorem 1.

Suppose that $\mathbb{E}[X^{2}]<\infty$. Then there is a $\mathcal{G}$-measurable random variable $Y$ satisfying $\mathbb{E}[Y^{2}]<\infty$ and equation (1) is satisfied for all $G\in\mathcal{G}$.

###### Proof.

Consider the norm $\|Y\|_{2}\equiv\mathbb{E}[Y^{2}]^{1/2}$ on the vector space $V=L^{2}(\Omega,\mathcal{F},\mathbb{P})$ of real valued random variables $Y$ satisfying $\mathbb{E}[Y^{2}]<\infty$ (up to $\mathbb{P}$ almost everywhere equivalence). This is given by the following inner product  $\langle Y_{1},Y_{2}\rangle\equiv\mathbb{E}[Y_{1}Y_{2}].$

As $L^{p}$-spaces are complete  , this makes $V$ into a Hilbert space  (see also, $L^{2}$-spaces are Hilbert spaces (http://planetmath.org/L2SpacesAreHilbertSpaces)). Then, $U\equiv L^{2}(\Omega,\mathcal{G},\mathbb{P})$ is a complete, and hence closed, subspace  of $V$.

By the existence of orthogonal projections (http://planetmath.org/ProjectionsAndClosedSubspaces) onto closed subspaces of Hilbert spaces, there is an orthogonal projection $\pi\colon V\rightarrow U$. In particular, $\langle\pi Y-Y,Z\rangle=0$ for all $Y\in V$ and $Z\in U$. Setting $Y=\pi X$ gives

 $\mathbb{E}[1_{G}Y]-\mathbb{E}[1_{G}X]=\langle 1_{G},\pi X-X\rangle=0$

as required. ∎

We can now prove the existence of conditional expectations of nonnegative random variables. Note that here there are no integrability conditions on $X$.

###### Theorem 2.

Let $X$ be a nonnegative random variable taking values in $\mathbb{R}\cup\{\infty\}$. Then, there exists a nonnegative $\mathcal{G}$-measurable random variable $Y$ taking values in $\mathbb{R}\cup\{\infty\}$ and satisfying (1) for all $G\in\mathcal{G}$. Furthermore, $Y$ is uniquely defined $\mathbb{P}$-almost everywhere (http://planetmath.org/AlmostSurely).

###### Proof.

First, let $X_{n}=\min(n,X)$. As this is bounded  , theorem 1 says that the conditional expectations $Y_{n}=\mathbb{E}[Y_{n}\mid\mathcal{G}]$ exist. Furthermore, as $X_{0}=0$, we may take $Y_{0}=0$. For any $n$, setting $G=\{Y_{n+1} gives

 $\mathbb{E}[1_{G}(Y_{n}-Y_{n+1})]=\mathbb{E}[1_{G}(X_{n}-X_{n+1})]\leq 0.$

So $1_{G}(Y_{n}-Y_{n+1})$ is a nonnegative random variable with nonpositive expectation, hence is almost surely equal to zero. Therefore, $Y_{n+1}\geq Y_{n}$ (almost surely) and, by replacing $Y_{n}$ with the maximum of $Y_{1},\ldots\,Y_{n}$ we may suppose that $(Y_{n})$ is an increasing sequence of random variables. Setting $Y=\sup_{n}Y_{n}$, the monotone convergence theorem  gives

 $\mathbb{E}[1_{G}Y]=\lim_{n\rightarrow\infty}\mathbb{E}[1_{G}Y_{n}]=\lim_{n% \rightarrow\infty}\mathbb{E}[1_{G}X_{n}]=\mathbb{E}[1_{G}X]$

as required.

Finally, suppose that $\tilde{Y}$ is also a nonnegative $\mathcal{G}$-measurable random variable satisfying (1). For any $x\in\mathbb{R}$, setting $G=\{\tilde{Y}>Y,x>Y\}$ then $1_{G}Y$ is bounded and,

 $\mathbb{E}[1_{G}(\tilde{Y}-Y)]=\mathbb{E}[1_{G}X]-\mathbb{E}[1_{G}X]=0$

showing that $\mathbb{P}(G)=0$. Letting $x$ increase to infinity gives $\tilde{Y}\leq Y$ (almost surely) and, similarly, $Y\leq\tilde{Y}$ so that $Y=\tilde{Y}$ almost surely. ∎

Finally, we show existence of the conditional expectation of every random variable $X$ satisfying $\mathbb{E}[|X|\mid\mathcal{G}]<\infty$ almost surely. Note, in particular, that this is satisfied whenever $X$ is integrable, as

 $\mathbb{E}[\mathbb{E}[|X|\mid\mathcal{G}]]=\mathbb{E}[|X|]<\infty.$
###### Theorem 3.

Let $X$ be a random variable such that $\mathbb{E}[|X|\mid\mathcal{G}]<\infty$ almost surely. Then, there exists a $\mathcal{G}$-measurable random variable $Y$ such that $\mathbb{E}[1_{G}|Y|]<\infty$ and (1) is satisfied for every $G\in\mathcal{G}$ with $\mathbb{E}[1_{G}|X|]<\infty$.

Furthermore, $Y$ is uniquely defined up to $\mathbb{P}$-a.e. equivalence.

###### Proof.

The positive and negative parts $X_{+},X_{-}$ of $X$ satisfy

 $\mathbb{E}[X_{+}\mid\mathcal{G}]+\mathbb{E}[X_{-}\mid\mathcal{G}]=\mathbb{E}[|% X|\mid\mathcal{G}]<\infty$

almost surely. We can therefore set $Y_{\pm}\equiv\mathbb{E}[X_{\pm}\mid\mathcal{G}]$ and $Y=Y_{+}-Y_{-}$.

If $G\in\mathcal{G}$ satisfies $\mathbb{E}[1_{G}|X|]<\infty$ then $\mathbb{E}[1_{G}Y_{\pm}]=\mathbb{E}[1_{G}X_{\pm}]<\infty$, so $\mathbb{E}[1_{G}|Y|]<\infty$ and,

 $\mathbb{E}[1_{G}Y]=\mathbb{E}[1_{G}Y_{+}]-\mathbb{E}[1_{G}Y_{-}]=\mathbb{E}[1_% {G}X_{+}]-\mathbb{E}[1_{G}X_{-}]=\mathbb{E}[1_{G}X]$

as required.

Finally, suppose that $\tilde{Y}$ satisfies the same conditions as $Y$. For any $x\geq 0$ set $G=\{Y_{+}+Y_{-}\leq x,\tilde{Y}>Y\}\in\mathcal{G}$. Then,

 $\mathbb{E}[1_{G}|X|]=\mathbb{E}[1_{G}(Y_{+}+Y_{-})]\leq x<\infty.$

So, $\mathbb{E}[1_{G}|Y|]$ and $\mathbb{E}[1_{G}|\tilde{Y}|]$ are finite, hence (1) gives

 $\mathbb{E}[1_{G}(\tilde{Y}-Y)]=\mathbb{E}[1_{G}X]-\mathbb{E}[1_{G}X]=0.$

So $\mathbb{P}(G)=0$ and, letting $x$ increase to infinity, $\tilde{Y}\leq Y$ almost surely. Similarly, $Y\leq\tilde{Y}$ and therefore $\tilde{Y}=Y$ almost surely. ∎

Title existence of the conditional expectation ExistenceOfTheConditionalExpectation 2013-03-22 18:39:28 2013-03-22 18:39:28 gel (22282) gel (22282) 5 gel (22282) Theorem msc 60A10